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Old 2007-12-19, 13:31   #1
Jean Penné
 
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Default Bases 2 & 4 reservations/statuses/primes

Hello Gary,

Congrats for organizing this search!
I am working alone on the Riesel base 4 candidates for a while, so I wish to reserve all the divisible by 3 k's candidates, up to n = 131072 base 4 for now...

The remaining k's I wish to reserve are :

9519, 13854, 14361, 16734, 19401 and 20049, which are marked as available on your reservation page. I already eliminated all the others...

Regards,
Jean
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Old 2007-12-19, 14:34   #2
gd_barnes
 
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Quote:
Originally Posted by Jean Penné View Post
Hello Gary,

Congrats for organizing this search!
I am working alone on the Riesel base 4 candidates for a while, so I wish to reserve all the divisible by 3 k's candidates, up to n = 131072 base 4 for now...

The remaining k's I wish to reserve are :

9519, 13854, 14361, 16734, 19401 and 20049, which are marked as available on your reservation page. I already eliminated all the others...

Regards,
Jean

Hi Jean,

Thank you and welcome to the effort! It's nice to have you on board. The k's are available and I will show them reserved by you today. It looks like you're now the official base 4 person; both Riesel and Sierpinski!

If you have the time, come vote in our poll in the lounge here.


Gary
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Old 2007-12-20, 03:32   #3
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Originally Posted by gd_barnes View Post
Hi Jean,
...
It looks like you're now the official base 4 person; both Riesel and Sierpinski!
...


Gary
I'm still around on the Sierpinski side of things....granted I'm really slow...but I'm still around. people are free to reserve any range they want but since no one has stepped forward I keep slowly going forward with it...

64494 is at 1770506 and still no prime
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Old 2007-12-20, 03:54   #4
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Originally Posted by tcadigan View Post
I'm still around on the Sierpinski side of things....granted I'm really slow...but I'm still around. people are free to reserve any range they want but since no one has stepped forward I keep slowly going forward with it...

64494 is at 1770506 and still no prime

Thanks for the update Tcadigan. I'll put you down for testing completion of n=885.2K base 4. I had previously noticed that you were searching one k for Sierp base 4 so sorry about the omission there.

If you'd like to pick up a smaller search effort that also LLR's as fast as base 2, we've got 55 k's for Sierp base 16, all tested to just n=25K that are ready for testing. At some future time, we also want to start on Riesel base 16 from the beginning.


Thanks,
Gary
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Old 2008-01-08, 06:27   #5
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Default Riesel/Sierp base 2/4 reservations/statuses/primes

Quote:
Originally Posted by Jean Penné View Post

About the second question, when the base is a power of two, there may be MOB k values which are also power of two, and then the candidates are not only Generalized Fermat Numbers, but very Fermat Numbers, so, I think these k's should be excluded (It is conjectured that there are a finite number of Fermat primes, and most of the mathematicians believe that F4 = 65537 is the largest one...).
As I remarked in another thread, proving that F4 is the largest Fermat prime is equivalent to proving that 65536 is an even Sierpinski number...
But, if so, the covering set would be infinite, because it is well known that Fermat numbers are pairwise coprime...

Regards,
Jean
(Edit note: MOB = multiples of base)

Robert, Citrix, Axn, Geoff, Masser, or other people with intimate knowledge of the math's behind the conjectures, here is what I would propose (if it has not been proposed already) that Jean agrees with:

In order to prove the Sierpinski conjecture for any base, all Generalized Fermat #'s as well as very Fermat #'s, i.e. any form that reduces to 2^n+1, should be excluded from those conjectures.

In a nutshell, here is what I'm working towards as determining the Riesel/Sierpinski conjecture proofs on the "Conjectures 'R Us" web pages:


1. All generalized Fermat #'s (18*18^n+1) and very Fermat #'s (65536*4^n+1 or 65536*16^n+1) will be excluded from the conjectures.

2. Any k that obtains a full covering set in any manner from ALGEBRAIC factors will be excluded. In many instances, this includes k's where there is a partial covering set of numeric factors (or a single numeric factor) and a partial covering set of algebraic factors that combine to make a full covering set.

3. All k below the lowest k found to have a NUMERIC covering set must have a prime including multiples of the base (MOB) but excluding the conditions in #1 and #2 above. I will add MOB and exclude GFn's in the near future on the pages. There are very few MOB if GFNs are excluded.

4. All n must be >= 1.


All input and opinions are welcome.

Citrix, if you think this is misplaced, feel free to move it around somewhere.


Thanks,
Gary

Last fiddled with by gd_barnes on 2008-01-08 at 06:32
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Old 2008-01-08, 20:33   #6
Jean Penné
 
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Default Liskovets-Gallot numbers are beautiful for us!

Hi,

On the 23 May 2006, Citrix warned us, in the Sierpinski base 4 thread, about this problem :

http://www.primepuzzles.net/problems/prob_036.htm

To be short, the Liskovets assertion is :

There are some k values such that k*2^n+1 is composite for all n values of certain fixed parity, and some k values such that k*2^n-1 is composite for all n values of certain fixed parity.

It is almost evident that these k values must be searched only amongst the multiples of 3 (the assertion is trivial if 3 does not divide k) :

If k == 1 mod 3, then 3 | k*2^n-1 if n is even, and 3 | k*2^n+1 if n is odd.
If k == 2 mod 3, then 3 | k*2^n+1 if n is even, and 3 | k*2^n-1 if n is odd.

Almost immediately after, Yves Gallot discovered the first four Liskovets-Gallot numbers ever produced :

k*2^n+1=composite for all n=even: k=66741
k*2^n+1=composite for all n=odd: k=95283
k*2^n-1=composite for all n=even: k=39939
k*2^n-1=composite for all n=odd: k=172677

And Yves said that "I conjecture that 66741, 95283, 39939, ... and 172677 are the smallest solutions for the forms - having no algebraic factorization (such as 4*2^n-1 or 9*2^-1) - but I can't prove it."

For several reasons, I think it would be interesting for us to coordinate the search in order to prove these four conjectures :

1) They involve only k values that are multiples of 3, so the success will no more be depending of the SoB or Rieselsieve one.

2) For the n even Sierpinski case, only k = 23451 and k = 60849 are remaining, with n up to more than 1,900,000 that is to say there are only two big primes to found, then the conjecture is proven!

3) For the n even Riesel (third line above) there are only four k values remaining : 9519, 14361, 19401 and 20049, although the search is only at the beginning!

4) For the two remaining n odd Sierpinski / Riesel (which can be tranlated as
base 4, k even, and doubling the Gallot values : 190566 for k*4^n+1, 345354 for k*4^n-1) I began to explore the problem, by eliminating all k's yielding a prime for n < 4096, eliminating the perfect square k values for Riesel, eliminating the MOB that are redondant, etc...

Finally, there were 42 k values remaining for +1, 114 for -1, and after sieving rapidly with NewPGen, and LLRing up to ~32K, I have now 21 values remaining for +1 and 37 values for -1.

I would be happy to know your opinion about all that...
Regards,
Jean

Last fiddled with by LaurV on 2019-06-14 at 16:51 Reason: fixed few typos
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Old 2008-01-10, 19:15   #7
gd_barnes
 
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Quote:
Originally Posted by Jean Penné View Post
Hi,

On the 23 May 2006, Citrix warned us, in the Sierpinski base 4 thread, about this problem :

http://www.primepuzzles.net/problems/prob_036.htm

To be short, the Liskovets assertion is :

There are some k values such that k*2^n+1 is composite for all n values of certain fixed parity, and some k values such that k*2^n-1 is composite for all n values of certain fixed parity.

It is almost evident that these k values must be searched only amongst the multiples of 3 (the assertion is trivial if 3 does not divide k) :

If k == 1 mod 3, then 3 | k*2^n-1 if n is even, and 3 | k*2^n+1 if n is odd.
If k == 2 mod 3, then 3 | k*2^n+1 if n is even, and 3 | k*2^n-1 if n is odd.

Almost immediately after, Yves Gallot discovered the firt four Liskovets-Gallot numbers ever produced :

k*2^n+1=composite for all n=even: k=66741
k*2^n+1=composite for all n=odd: k=95283
k*2^n-1=composite for all n=even: k=39939
k*2^n-1=composite for all n=odd: k=172677

And Yves said that "I conjecture that 66741, 95283, 39939, ... and 172677 are the smallest solutions for the forms - having no algebraic factorization (such as 4*2^n-1 or 9*2^-1) - but I can't prove it."

For several reasons, I think it would be interesting for us to coordinate the search in order to prove these four conjectures :

1) They involve only k values that are multiples of 3, so the success will no more be depending of the SoB or Rieselsieve one.

2) For the n even Sierpinski case, only k = 23451 and k = 60849 are remaining, with n up to more than 1,900,000 that is to say there are only two big primes to found, then the conjecture is proven!

3) For the n even Riesel (third line above) there are only four k values remaining : 9519, 14361, 19401 and 20049, although the search is only at the beginning!

4) For the two remaining n odd Sierpinski / Riesel (which can be tranlated as
base 4, k even, and doubling the Gallot values : 190566 for k*4^n+1, 345354 for k*4^n-1) I began to explore the problem, by eliminating all k's yielding a prime for n < 4096, eliminating the perfect square k values for Riesel, eliminating the MOB that are redondant, etc...

Finally, there were 42 k values remaining for +1, 114 for -1, and after sieving rapidly with NewPGen, and LLRing up to ~32K, I have now 21 values remaining for +1 and 37 values for -1.

I would be happy to know your opinion about all that...
Regards,
Jean

I responded in the "Conjecture 'R Us" thread to this.


Gary
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Old 2008-01-10, 19:29   #8
Jean Penné
 
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Originally Posted by kar_bon View Post
i read about your post but no time yet to work with.
it's quite an interesting conjecture to prove and i'm interested in the Riesel odd/even side of this problem. it's worth another extra data page for the RPS site.
will do this for the Riesel power-2 bases for CRUS too next time.
i think i take some searching on 9519, 14361, 19401 and 20049. which ranges you searched yet? any findings?
karsten
For now, my progress on these k's follows (exponents n are in base 2) :

k = 9519 is up to n = 427816, no prime.
k = 14361 is up to n = 318510, no prime.
k = 19401 is up to n = 262578, no prime.
k = 20049 is up to n = 265144, no prime.

I thought to test these k's up to 524288 base 2 if no prime found below...
However, to day, only 9519 is running... If you have more available computing power, you may continue on the last three k's, but if so, let me know about your progress.

Regards,
Jean
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Old 2008-01-10, 21:42   #9
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Quote:
Originally Posted by Jean Penné View Post
For now, my progress on these k's follows (exponents n are in base 2) :

k = 9519 is up to n = 427816, no prime.
k = 14361 is up to n = 318510, no prime.
k = 19401 is up to n = 262578, no prime.
k = 20049 is up to n = 265144, no prime.

I thought to test these k's up to 524288 base 2 if no prime found below...
However, to day, only 9519 is running... If you have more available computing power, you may continue on the last three k's, but if so, let me know about your progress.

Regards,
Jean
OK, I will make note of the test limits and unreserve k=13854, 16734, and 19464. Note that the last one is currently only on the base 16 page because it's a MOB base 4.

Here is what I show for the test limits for the k's that you're unreserving:

k=13854 and 16734; I show no work so they are still at n=100K base 4.
k=19464 per your status report of 1/7/08 is at n=137112 base 2.

Is that correct?

I'll coordinate getting the 3 unreserved k's tested. If they aren't reserved in the new few days, I'll reserve and test them.

Since you did some testing on k=19464, do you have a sieve file for that one or for any of the k's? If so, I will post them on the website.


Gary
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Old 2008-01-10, 21:54   #10
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got sieve-files from jean for 14361, 19401 and 20049 and sieving further!
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Old 2008-01-10, 23:21   #11
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Quote:
Originally Posted by kar_bon View Post
got sieve-files from jean for 14361, 19401 and 20049 and sieving further!
I messed up originally here by not following the entire thread. I'll straighten out the reservations shortly. Sorry about my confusion.

You can ignore my last post.

Edit: It's all correct now. You might check the Riesel base 4 and base 16 reservations to make sure I have it correct.


Gary

Last fiddled with by gd_barnes on 2008-01-11 at 03:08 Reason: Figured it out; will straight out res.
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